Problem: Let $f(x)=x^4-2x^3$. For what value of $x$ does $f$ have a relative minimum ? Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{2}{3}$ (Choice B) B $0$ (Choice C) C $\dfrac{3}{2}$ (Choice D) D $2$
Explanation: We can find the relative extrema (i.e. minima and maxima) of $f$ by looking for the intervals where its derivative $f'$ is positive/negative. A function can only change its direction from increasing to decreasing and vice versa between its critical points and the points where the function itself is undefined. The derivative of $f$ is $f'(x)=2x^2(2x-3)$. $f'(x)=0$ for $x=0,\dfrac{3}{2}$. Since $f'$ is a polynomial, it's defined for all real numbers. Therefore, our critical points are $x=0$ and $x=\dfrac{3}{2}$. $f$ is defined for all real numbers so we only need to consider the critical points. Our critical points divide the number line into three intervals: $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $(-\infty,0)$ $(0,\frac{3}{2})$ $\frac{3}{2}$ $(\frac{3}{2},\infty)$ Let's evaluate $f'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $f'(x)$ Verdict $(-\infty,0)$ $x=-1$ $f'(-1)=-10<0$ $f$ is decreasing $\searrow$ $(0,\dfrac{3}{2})$ $x=\dfrac{1}{2}$ $f'\left(\dfrac12\right)=-1<0$ $f$ is decreasing $\searrow$ $(\dfrac{3}{2},\infty)$ $x=2$ $f'(2)=8>0$ $f$ is increasing $\nearrow$ Now let's look at the critical points: $x$ Before After Verdict $0$ $\searrow$ $\searrow$ Not an extremum $\dfrac32$ $\searrow$ $\nearrow$ Minimum Now we can see that $f$ has a relative minimum at $x=\dfrac{3}{2}$.